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Given a function $f \colon \mathbb{R^n} \to E$, where $E$ is a locally convex Hausdorff TVS, we say that $f$ is differentiable at $x \in \mathbb{R^n}$ if there exists some $Df(x) \in \mathcal{L}(\mathbb R ^n, E)$ such that $$ f(x+h) = f(x) + Df(x)h + o(|h|). $$ Here $g = o(|h|)$ if $\lim_{h\to0} \frac{g(h)}{|h|} = 0$ as a limit in $E$.

Given such a function, we have for all $\psi \in E'$ that $$ \psi \circ f(x+h) = \psi \circ f(x) + \psi \circ Df(x)h + o(|h|). $$ (Note the slight abuse of little-o notation; the first one is $E$-valued while the second is $\mathbb{R}$-valued.) Since $\psi \circ Df(x) \in \mathcal{L}(\mathbb R ^n,\mathbb R)$, this means $\psi \circ f$ is differentiable in the classical sense.

Question 1: Does the reverse hold, i.e. if $\psi \circ f$ is diffentiable at $x$ for all $\psi \in E'$, is $f$ itself then differentiable at $x$? NO, see a nice counterexample given here.

Question 2: What if we look at differentiablility on an open $U$ instead of one point $x$?

mixotrov
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  • Could you clarify why $g=o(|h|) \iff \lim_{h \to 0} \frac{g(h)}{|h|} $ would not work in a general TVS? – mixotrov Mar 09 '23 at 16:14
  • I am not saying it does not work. Just a little addition to your posting -to clarify things- would be nice. There is a paper on differentiation on vector spaces that considers all types of questions, notions of differentiation, and examples that my be of help yo you. The original version (in russian) is in the link I am sending; the English translation requires some library privileges to access The London Mathematical Society journal. – Mittens Mar 09 '23 at 16:28
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    If you ask only about a single point $x$, then the answer is no even when $E$ is a Hilbert space, as I mention here. – peek-a-boo Mar 09 '23 at 16:57
  • Thank you for the counterexample! I'll update the question. – mixotrov Mar 10 '23 at 10:04
  • in the complex case with differentiability on an open set, I’ll say that in the Banach space context, it is true (as I mentioned in the linked answer), i.e on an open set, weak-holomorphy is equivalent to strong holomorphy. The proof relies on Hahn-Banach and Cauchy’s integral formula in the scalar case. Now, I’m rusty with my functional analysis, but if Hahn-Banach holds more generally for Frechet spaces, then I guess you could try to adapt the Banach space proof here as well (so I wouldn’t be surprised if true in C). In the real case though, this seems highly unlikely, – peek-a-boo Mar 10 '23 at 20:56

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